Suppose we have a matrix $$ M=\left( \begin{array}{ccc} 0 & -\Delta \omega & 0 \\ \Delta \omega & -\gamma & \bar{\omega } \\ 0 & -\bar{\omega } & -\gamma \\ \end{array} \right) $$ with $\lambda, \bar{\omega}, \Delta \omega \in \mathbb{R}$. It is perhaps worth noting that the matrix can equivalently be defined as acting on a vector $v$ to produce a vector $Mv$ given by $$ Mv = v \times \left( \bar{\omega}, 0, -\Delta\omega \right) - \gamma \left(1 - \hat{x} \hat{x}^T\right)v $$ where $\times$ is the cross product and $\hat{x} = \left(1,0,0\right)$.
If a vector $v(t)$ is governed by the equation $$ \dot{v}=M v $$ we have the solution $$ v(t) = e^{M(t-t_o)}v_o $$ for some initial condition $v_o = v(t_o)$. Practically the solution can be computed by determining the eigenvalues and eigenvectors of $M$.
Now suppose that $\Delta \omega$ is normally distributed about zero with a standard deviation $\sigma_{\Delta \omega}$. My question is: does the expectation value of $v(t)$ admit a relatively simple solution, say, in the form of $$ \langle v(t) \rangle = e^{M'(t-t_o)}v_o $$ for some matrix $M'$ that is a function of $\sigma_{\Delta \omega}$?
So far I can see that we have
$$ \langle v(t) \rangle = \sum_{n=0}^\infty \frac{(t-t_o)^n \langle M^n \rangle}{n!} v_o $$
but it is not clear to me if the infinite sum simplifies, or if the special form of $M$ enjoys any simplifications.